Prove that $\text{Frac}(\mathbb{Z}[\sqrt{d}]) \cong \mathbb{Q}(\sqrt{d})$

Question

I'm trying to prove that the field of fractions for $\mathbb{Z}[\sqrt{d}]$ is isomorphic to $\mathbb{Q}(\sqrt{d})$.
I know exactly how to go about this problem, but I'm just confused as to how to define the function.
I tried the following:
"Define a function $\phi: \text{Frac}(\mathbb{Z}[\sqrt{d}]) \to \mathbb{Q}(\sqrt{d})$ by $\phi(\frac{p+r\sqrt{d}}{s+t\sqrt{d}})=\frac{p}{s}+\frac{r}{t}\sqrt{d}$."
This function gave me some pretty ugly irreducible computations in terms of proving that it respected addition.
Is there another function I should be using?

Answer 1

Here is another take.

The field of fractions for $\mathbb{Z}[\sqrt{d}]$ is the smallest subfield $K$ of $\mathbb C$ that contains $\mathbb{Z}[\sqrt{d}]$.

Now $\mathbb Z \subseteq \mathbb{Z}[\sqrt{d}]$ implies $\mathbb Q \subseteq K$ and so $\mathbb{Q}(\sqrt{d}) \subseteq K$.

Since $\mathbb{Q}(\sqrt{d})$ is a field containing $\mathbb{Z}[\sqrt{d}]$, it must contain $K$, that is $\mathbb{Q}(\sqrt{d}) \supseteq K$.

Therefore, $K = \mathbb{Q}(\sqrt{d})$.